TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-12

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Every nontrivial extensional property of Turing-machine languages is undecidable

Statement

Let $P$ be a nontrivial extensional property of recognizable languages. Then its index set $I_{P} = {⌜ M ⌝ : L (M) \in P}$ is undecidable.

Facts & Assumptions

Given: A nontrivial extensional property $P$ and its index set $I_{P}$ .

[L1]

For such a property, A nontrivial extensional property admits a uniform witness machine construction gives a recognizable language $W$ opposite to $\emptyset$ with respect to $P$ and a total computable map $g$ such that $x \in A_{T M}$ implies $L (N_{x}) = W$ while $x \notin A_{T M}$ implies $L (N_{x}) = \emptyset$ .

[L2]

The language $A_{T M}$ is undecidable, by The Turing-machine acceptance problem is undecidable.

Proof

technique · contradiction

1.1

Assume for contradiction that $I_{P}$ is decidable. Fix the map $g$ and the language $W$ from [L1]. Because $W$ and $\emptyset$ lie on opposite sides of $P$ , exactly one of the two conditions $W \in P$ and $\emptyset \in P$ holds.

assume-contraL1given

2.1

On input $x$ , compute $g (x) = ⌜ N_{x} ⌝$ and run the assumed decider for $I_{P}$ on that code. If $W \in P$ , answer exactly as the decider answers; if $\emptyset \in P$ , flip the decider's answer. By [L1], this procedure accepts exactly those $x$ in $A_{T M}$ , so it decides $A_{T M}$ .

L1step 1.1construct

3.1

Step 2.1 contradicts [L2]. Therefore the assumption in step 1.1 was false, and $I_{P}$ is undecidable.

L2step 2.1contradictiondischarge-contradiction: a decider for the index set would decide A_TM∎

Depends on

Used by

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. G. Rice, Classes of Recursively Enumerable Sets and Their Decision Problems (standard reference, not scraped)
EECS 376 Course Notes, Part 6: Computability (standard reference, not scraped)